Exam 2A Solutions

# Exam 2A Solutions - &amp;amp e&amp;quot m y = sin 3 cos 3 3 5(8&amp;quot 1 2 \$&amp;amp =&amp;quot 1 2 \$&amp;amp h 1 2

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Mt100 - Calculus I Exam 2 Solutions Variation: A 1 . (8) a. 23.6 " 0.0 10 " 0 = 2.36 miles/minute b. 23.1 " 17.4 8 " 4 = 1.425 miles/minute c. Negative, since the distance is increasing at a decreasing rate. 2. f ( x ) (in black ), " f ( ) (in red ), " " ( ) (in green ).

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3 . (8) a. D (1) = 70 means that the train traveled 70 miles in the first hour. b. " D (2) = 40 means that 2 hours after leaving, the train is traveling at 40 miles per hour. 4 (28) " f ( x ) = 8 7 # 28 3 + 8 2 # 56 # 9 " g ( z ) = 4 3 sin + 4 cos c. " h ( # ) = 2 + 1 \$ % ( ) e * + 2 2 + 1 \$ % ( ) 2 d. " k ( ) = 20 10 + # \$ % ( 19 10 9 + 1 # \$
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Unformatted text preview: % &amp;amp; ( e. &amp;quot; m ( y ) = sin 3 ( ) # cos 3 ( ) # 3 5 (8) &amp;quot; 1 2 # \$ % % &amp;amp; ( ( = &amp;quot; 1 2 # \$ % % &amp;amp; ( ( h 1 2 # \$ % % &amp;amp; ( ( + f 1 2 # \$ % % &amp;amp; ( ( &amp;quot; 1 2 # \$ % % &amp;amp; ( ( = ) 7 4 + * 1 ( ) ) 2 = * 2 &amp;quot; w ( # 1) = &amp;quot; g h ( # 1) ( ) \$ &amp;quot; ( # 1) = &amp;quot; 1 ( ) \$ ( # 1) = ( # 4) \$ ( # 1) = 4...
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## This note was uploaded on 06/04/2008 for the course MT 100 taught by Professor Keane during the Fall '07 term at BC.

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Exam 2A Solutions - &amp;amp e&amp;quot m y = sin 3 cos 3 3 5(8&amp;quot 1 2 \$&amp;amp =&amp;quot 1 2 \$&amp;amp h 1 2

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